436. Find Right Interval

Given a set of intervals, for each of the interval i, check if there exists an interval j whose start point is bigger than or equal to the end point of the interval i, which can be called that j is on the "right" of i.

For any interval i, you need to store the minimum interval j's index, which means that the interval j has the minimum start point to build the "right" relationship for interval i. If the interval j doesn't exist, store -1 for the interval i. Finally, you need output the stored value of each interval as an array.

Note:

1. You may assume the interval's end point is always bigger than its start point.
2. You may assume none of these intervals have the same srt point.

Example 1:

Input: [ [1,2] ]

Output: [-1]

Explanation: There is only one interval in the collection, so it outputs -1.

Example 2:

Input: [ [3,4], [2,3], [1,2] ]

Output: [-1, 0, 1]

Explanation: There is no satisfied "right" interval for [3,4].
For [2,3], the interval [3,4] has minimum-"right" start point;
For [1,2], the interval [2,3] has minimum-"right" start point.

Example 3:

Input: [ [1,4], [2,3], [3,4] ]

Output: [-1, 2, -1]

Explanation: There is no satisfied "right" interval for [1,4] and [3,4].
For [2,3], the interval [3,4] has minimum-"right" start point.

class Solution {
    public int[] findRightInterval(int[][] intervals) {
        int n = intervals.length;
        int[] res = new int[n];
        Arrays.fill(res, -1);
        
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < n; j++) {
                if(intervals[i][1] <= intervals[j][0]) {
                    if(res[i] == -1) {
                       res[i] = j; 
                    }
                    else if(intervals[j][0] < intervals[res[i]][0]) res[i] = j;
                }
            }
        }
        return res;
    }
}

bruteforce，太brute了

class Solution {
    public int[] findRightInterval(int[][] intervals) {
        int n = intervals.length;
        int[] res = new int[n];
        TreeMap<Integer, Integer> map = new TreeMap();
        for(int i = 0; i < n; i++) map.put(intervals[i][0], i);
        
        for(int i = 0; i < n; i++) {
            Integer ind = map.ceilingKey(intervals[i][1]);
            res[i] = ind == null ? -1 : map.get(ind);
        }
        return res;
    }
}

题目说了所有start point都不一样，而且我们要找的是每个的最小start point，利用这个性质可以搞一个treemap，key是start point，value是index

存完之后就遍历，找有没有对应这个interval end point的ceilingKey（大于等于），有就get到对应的index存入res中。

好赖啊

标签：Right,point,int,Interval,interval,start,intervals,right,436
来源： https://www.cnblogs.com/wentiliangkaihua/p/13575113.html